/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Students may choose between a 3-... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Students may choose between a 3-semester-hour course in physics without labs and a 4-semester-hour course with labs. The final written examination is the same for each section. If 12 students in the section with labs made an average examination grade of 84 with a standard deviation of \(4,\) and 18 students in the section without labs made an average: grade of 77 with a standard deviation of \(6,\) find a \(99 \%\). confidence interval for the difference between the average grades for the two courses. Assume the populations to be approximately normally distributed with equal variances.

Short Answer

Expert verified
The 99% confidence interval for the difference between the average grades for the two courses can be computed following these steps.

Step by step solution

01

Determine sample sizes, means, and standard deviations

First, recognize the given values, the sample sizes \(n_1\) and \(n_2\), the means \(\bar{x}_1\) and \(\bar{x}_2\), and the standard deviations \(s_1\) and \(s_2\) for labs and no labs classes respectively. Here, \(n_1 = 12\), \(\bar{x}_1 = 84\), \(s_1 = 4\), \(n_2 = 18\), \(\bar{x}_2 = 77\), \(s_2 = 6\).
02

Find the standard error of the difference

The standard error for the difference of two means can be found using the formula: \[SE = \sqrt{ \frac{{s_1}^{2}}{n_1} + \frac{{s_2}^{2}}{n_2} }\] Put the known values into the formula: \[SE = \sqrt{ \frac{{4}^{2}}{12} + \frac{{6}^{2}}{18} }\]
03

Find the 99% confidence interval

For a 99% confidence level, the z-score (z) is approximately 2.576 (usually found in statistical tables). Use the formula for the confidence interval for the difference of two means: \[CI = ((\bar{x}_1 - \bar{x}_2) - z * SE, (\bar{x}_1 - \bar{x}_2) + z * SE)\] Put the calculated standard error and the known values into the formula: \[CI = ((84 - 77) - 2.576 * SE, (84 - 77) + 2.576 * SE)\] to get the desired confidence interval.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Error
The standard error (SE) is crucial in statistics as it measures the precision with which a sample mean represents the population mean. Essentially, it gauges the variation of the sample means around the true population mean. The lower the standard error, the more accurate the estimate.

In the context of comparing two means, the SE is modified to account for the variability within both samples. It's calculated using the formula: \[ SE = \sqrt{ \frac{{s_1}^{2}}{n_1} + \frac{{s_2}^{2}}{n_2} } \]where \(s_1\) and \(s_2\) are the standard deviations of the sample means, and \(n_1\) and \(n_2\) are the sample sizes. Knowing that the SE quantifies how much the difference between the sample means might vary, allows us to construct a range (interval) within which the actual difference between the population means likely falls.

For our exercise, calculating the SE is a step toward establishing how confident we can be about the difference in exam grades between students taking physics with labs versus without labs.
Difference of Two Means
The difference of two means is a statistical measure used to compare the averages from two distinct groups or samples. In the exercise, we're contrasting the average final examination grades for two groups of students: those who enrolled in a physics course with labs and those without.

When comparing two sample means, the actual point of interest is not the means themselves, but the difference between them. We denote this as \((\bar{x}_1 - \bar{x}_2)\). This difference has its own variability and distribution, which is where the standard error comes into play, providing a basis for how much this difference can fluctuate due to sampling error.

In the provided exercise, after identifying the sample means and their standard deviations, we then consider this difference under the assumption of normality, which allows us to employ the normal distribution's properties to construct a confidence interval around the difference.
Normal Distribution
A cornerstone of statistics is the normal distribution, a bell-shaped curve that is symmetric about the mean, depicting how values in a dataset are dispersed. Most values cluster around the mean, and as one moves away from the mean, the frequency of values decreases. It has many wonderful properties, one of which is that it's completely defined by two parameters: the mean and the standard deviation.

When considering the confidence interval for the difference of two means, we inexorably rely on the assumption that the distribution of these differences approximates, or is, normal. This assumption allows for the use of z-scores, which are standardized scores representing the number of standard deviations away from the mean a value lies.

Applying this to our exercise, we use the z-score corresponding to a 99% confidence level (approximately 2.576 for a two-tailed test) to indicate how far the difference of the sample means (our point estimate) should be adjusted to create an interval likely containing the true mean difference, given the normal distribution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A study was conducted to determine if a certain metal treatment has any effect on the amount of metal removed in a pickling operation. A random sample of 100 pieces was immersed in a bath for 24 hours without the treatment, yielding an average of 12.2 millimeters of metal removed and a sample standard deviation of 1.1 millimeters. A second sample of 200 pieces was exposed to the treatment, followed by the 24 -hour immersion in the bath, resulting in an average: removal of 9.1 millimeters of metal with a sample standard deviation of 0.9 millimeter. Compute a \(98 \%\) confidence interval estimate for the difference between the population means. Docs the treatment appear to reduce the mean amount of metal removed?

A type of thread is being studied for its tensile strength properties. Fifty pieces were tested under similar conditions and the results showed an average tensile strength of 78.3 kilograms and a standard deviation of 5.6 kilograms. Assuming a normal distribution of tensile strength, give a lower \(95 \%\) prediction interval on a single observed tensile strength value. In addition, give a lower \(95 \%\) tolerance limit that is exceeded by \(99 \%\) of the tensile strength values.

A random sample of 25 bottles of buffered aspirin contain, on average, \(325.05 \mathrm{mg}\) of aspirin with a standard deviation of \(0.5 \mathrm{mg}\). Find the \(95 \%\) tolerance limits that will contain \(90 \%\) of the aspirin contents for this brand of buffered aspirin. Assume that the aspirin content is normally distributed.

A certain supplier manufactures a type of rubber mat that is sold to automotive companies. In the application, the pieces of material must have certain hardness characteristics. Defective mats are occasionally discovered and rejected. The supplier claims that the proportion defective is \(0.05 .\) A challenge was made by one of the clients who purchased the product. Thus an experiment was conducted in which 400 mats are tested and 17 were found defective. (a) Compute a \(95 \%\) two-sided confidence interval on the proportion defective. (b) Compute an appropriate \(95 \%\) one-sided confidence interval on the proportion defective. (c) Interpret both of these in (a) and (b) and comment on the claim made by the supplier.

The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. (a) Construct a \(98 \%\) confidence interval for the mean height of all college students. (b) What can we assert with \(98 \%\) confidence about the possible size of our error it we estimate the mean height of all college students to be 174.5 centimeters?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.